Propane burns according to the following equation: C3H8+5O2-->3CO2+4H2O

Question

(a)How many grams of O2 are needed to react fully with 6.18 mol of propane?

(b)How many grams of CO2 can form from 0.576 mol oof propane?

(c)How many grams of water are produced by the combustion of 4.54 mol of propane?

Answer 1

C₃H₈ + 5O₂ ==> 3CO₂ + 4H₂O ... balanced equation

We will be using dimensional analysis to answer these questions.

molar mass O₂ = 32 g / mol

molar mass CO₂ = 44 g / mol

molar mass H₂O = 18 g / mol

(a) 6.18 mol C₃H₈ x 5 mol O₂ / 1 mol C₃H₈ x 32 g O₂ / mol = g O₂

(b) 0.576 mol C₃H₈ x 3 mol CO₂ / 1 mol C₃H₈ x 44 g CO₂ / mol = g CO₂

(3) 4.54 mol C₃H₈ x 4 mol H₂O / 1 mol C₃H₈ x 18 g H₂O / mol = g H₂O

Just do the math and you'll get the ansers